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根据一棵树的前序遍历与中序遍历构造二叉树。


注意:你可以假设树中没有重复的元素。


例如，给出前序遍历 preorder = [3,,"> 
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        <h1 class="title">leetcode105. 从前序与中序遍历序列构造二叉树</h1>
        <div class="stuff">
            <span>三月 14, 2020</span>
            

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            <h1 id="leetcode105-从前序与中序遍历序列构造二叉树"><a href="#leetcode105-从前序与中序遍历序列构造二叉树" class="headerlink" title="leetcode105. 从前序与中序遍历序列构造二叉树"></a>leetcode105. 从前序与中序遍历序列构造二叉树</h1><hr>
<blockquote>
<p>根据一棵树的前序遍历与中序遍历构造二叉树。</p>
</blockquote>
<blockquote>
<p>注意:<br>你可以假设树中没有重复的元素。</p>
</blockquote>
<blockquote>
<p>例如，给出<br>前序遍历 preorder = [3,9,20,15,7]<br>中序遍历 inorder = [9,3,15,20,7]<br>返回如下的二叉树：</p>
</blockquote>
<pre><code>  3
 / \
9  20
  /  \
 15   7</code></pre><p>先看这么一颗树:</p>
<pre><code>         1
   /        \
  2         3
 / \        / \
4   5     6   7
         /  \
        8    9</code></pre><p>前序遍历为:  [1,2,4,5,3,6,8,9,7]<br>中序遍历为:  [4,2,5,1,8,6,9,3,7]<br>这棵树的根节点为 <strong>1</strong>,能够由前序遍历的第一个元素知道.<br>然后看中序遍历, <strong>1</strong> 的左边,为左子树,<strong>1</strong> 的右边为右子树.以左子树 <strong>4,2,5</strong> 为例.<br>在前序遍历中,2,4,5和3,6,8,9,7分别为1的左右子树 . 遍历2,4,5第一个遍历到的是2,说明2为这棵树的根节点,然后再去中序遍历找,这么递归下去.<br>[1,==2,4,5==,==3,6,8,9,7==]<br>[==4,2,5==,1,==8,6,9,3,7==]</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> TreeNode <span class="title">buildTree</span><span class="params">(<span class="keyword">int</span>[] preorder, <span class="keyword">int</span>[] inorder)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(preorder.length==<span class="number">0</span>)<span class="keyword">return</span> <span class="keyword">null</span>;</span><br><span class="line">        <span class="keyword">return</span> build(preorder,inorder,<span class="number">0</span>,inorder.length-<span class="number">1</span>,<span class="number">0</span>);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> TreeNode <span class="title">build</span><span class="params">(<span class="keyword">int</span>[]preorder,<span class="keyword">int</span>[]inorder,<span class="keyword">int</span> is,<span class="keyword">int</span> ie,<span class="keyword">int</span> now)</span></span>&#123;</span><br><span class="line">    <span class="comment">//is和ie,表示inorder中的起始位置和结束位置,也就是这棵树的范围,now表示preorder中的根节点位置.</span></span><br><span class="line">        <span class="keyword">if</span>(is&gt;ie||is&lt;<span class="number">0</span>||ie&gt;inorder.length)<span class="keyword">return</span> <span class="keyword">null</span>;<span class="comment">//这个范围是我测试很多组数据试出来的</span></span><br><span class="line">        TreeNode node=<span class="keyword">new</span> TreeNode(preorder[now]);</span><br><span class="line">        <span class="keyword">if</span>(is==ie)<span class="keyword">return</span> node;</span><br><span class="line">        <span class="keyword">int</span> i=is;</span><br><span class="line">        <span class="keyword">for</span>(i=is;i&lt;ie;i++)&#123;<span class="comment">//在inorder内查找根节点的位置</span></span><br><span class="line">            <span class="keyword">if</span>(preorder[now]==inorder[i])<span class="keyword">break</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">//对于前序遍历来说,左子树的根节点是now+1,比如上面的preorder1后面是2</span></span><br><span class="line">        <span class="comment">//右子树的根节点需要用now再加上左子树的节点数量</span></span><br><span class="line">        node.left=build(preorder,inorder,is,i-<span class="number">1</span>,now+<span class="number">1</span>);</span><br><span class="line">        node.right=build(preorder,inorder,i+<span class="number">1</span>,ie,now+i-is+<span class="number">1</span>);</span><br><span class="line">        <span class="keyword">return</span> node;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p><strong>leetcode 47/100</strong></p>

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